electromagnetism - Does the power weakening of an electromagnetic transmission over distance depends on the beam's width?

Does the power weakening of an electromagnetic transmission over distance depends on the beam's width?

If I emit an omnidirectional electromagnetic transmission, its power weakens in the ratio of $1/R^2$, if I remember correctly.

My question is if I transmit a very narrow beam (say, 0.001 degrees, and "collect" the whole "transmission" at the receiver, so that nothing "escapes"), is it still true? Laser, for instance, doesn't suffer this kind of attenuation-scale (again, if I remember correctly...), due to its coherency.

Answer

Two parts to this answer - "near field" and "far field".

When you have an parabolic antenna, or in general any source that is not a point source, you can shape the beam to follow something other than a simple $1/r^2$ relationship - as you pointed out, a laser beam can be created in a way that you can "collect" all the energy (and assuming no absorption in the transmission medium).

This does not mean that the inverse square law is dead. In the limit where the source aperture is small compared to the distance, it once again comes into play. See this diagram (which shows "near field" at the top, and "far field" at the bottom):

Note that "source aperture" and "beam divergence" are closely related concepts, because of diffraction. This is described by the well known resolution equation:

$$\alpha = \frac{1.22\lambda}{d}$$

Which describes the angle at which the first null appears in the diffraction pattern for a circular aperture (Google Airy disk for more information. While that is usually considered an optical law, it applies to all electromagnetic radiation).

In short - you can try to trick the inverse square law, but over sufficiently long distances you can not beat it with a finite size receiver (as long as the receive aperture is larger than the diverged beam, you can in principle recover the energy; but note that if you only go as far out as the first minimum of the Airy disk, you will still lose "some" energy from the higher order fringes; in practice this means you will never recover all energy, even if you ignore attenuation in the medium).

So to address your specific question - even over short distances, you will lose a small amount of the energy due to diffraction effects; but when your receive dish is large enough to cover up to the first minimum, you will get a good percentage of the power - in fact a quick numerical integration of the intensity function

$$I(r) = \left(\frac{J_1(r)}{r}\right)^2$$

Gives the following results:

From 0 to 3.8317 = 0.414374
From 0 to 1000   = 0.424413

which suggests that about 2.3% of the power is lost outside of that first minimum. This is just approximate, of course...